是否可以将 HTML 输入类存储到 PHP 变量中？

Question

我有一个 html 表单和一些 php 以及一点点 javascript。该表单有两个输入标签。两个输入标签都有类属性。我想将类值“存储”在 PHP 变量中，以便在单击提交后回显。

我尝试将 javascript 与第一个 php 变量（$firstclass）集成，但即使作为警报（）也无法使其正常工作。我真的不想提醒类值，但认为这将有助于找到解决方案。

<form action="" method="post">

    <input type="text" name="input1" class="hidden_class_1">

    <input type="text" name="input2" class="hidden_class_2">

    <input type="submit" name="submit">

</form>

<?php

$firstclass = ""; //hidden_class_1

$secondclass = ""; //hidden_class_2

$firstclass = "<script type=\"application/javascript\">alert(('this.className').attr('class'))</script>";

$secondclass = ""; //ideally hidden_class_2

if(isset($_POST['submit'])){

    echo "<h2>First Input Class Value: ".$firstclass."</h2>";

    echo "<h2>Second Input Class Value: ".$secondclass."</h2>";

}

我希望输出如下；

第一个输入类值： hidden_class_1

第二个输入类值： hidden_class_2

Answer 1

“最简单”的方法是使用 AJAX/XHR 并将类发送到 PHP 脚本。<form id="ajaxform" action="path/to/script.php" method="post">&nbsp; <input type="text" name="input1" class="hidden_class_1">&nbsp; <input type="text" name="input2" class="hidden_class_2">&nbsp; <input type="submit" name="submit"></form>例如，使用 jQuery：const $form = $('#ajaxform');function onSuccess (response) {&nbsp; console.log('Successfully submitted the form');&nbsp; console.log('Server responded with', response);}function onFailure (jqXhr, status) {&nbsp; console.log('Ooops, something went wrong!');&nbsp; console.log('Server sent status code', status);}$form.on('submit', event => {&nbsp; event.preventDefault(); // suppress the reload&nbsp; const $input1 = $form.find('[name=input1]');&nbsp; const $input2 = $form.find('[name=input2]');&nbsp; $.ajax({&nbsp; &nbsp; method: $form.prop('method').toUpperCase(),&nbsp; &nbsp; url: $form.prop('action'),&nbsp; &nbsp; data: {&nbsp; &nbsp; &nbsp; input1Value: $input1.val(),&nbsp; &nbsp; &nbsp; input2Value: $input2.val(),&nbsp; &nbsp; &nbsp; input1Class: $input1.prop('className'),&nbsp; &nbsp; &nbsp; input2Class: $input2.prop('className')&nbsp; &nbsp; }&nbsp; }).&nbsp; done(onSuccess).&nbsp; fail(onFailure);});在您的 PHP 中，您将使用$_POST(或$_REQUEST) 来获取已发送的值：$input1_value = $_POST['input1Value'];$input2_value = $_POST['input2Value'];$input1_class = $_POST['input1Class'];$input2_class = $_POST['input2Class'];# do what you want with the variables请注意，您必须在onSuccess函数内部处理服务器的响应。通常，人们使用 JSON 对来自服务器的响应进行建模。您可以使用 PHP 的内置函数json_encode和json_decode函数。例如，您的 PHP 脚本可以回答：$input1_value = $_POST['input1Value'];$input2_value = $_POST['input2Value'];$input1_class = $_POST['input1Class'];$input2_class = $_POST['input2Class'];# do what you want to do with the variables, then$response = array(&nbsp; 'ok' => true,&nbsp; 'message' => 'PHP says "Thanks" for the information');header('Content-Type: application/json');echo json_encode($response);die;在onSuccess函数内部，您将例如：function onSuccess (response) {&nbsp; if (response.ok) {&nbsp; &nbsp; console.log('Submitted, and all values where OK');&nbsp; &nbsp; console.log(response.message);&nbsp; &nbsp; return; // opt-out early, no need for "else" keyword&nbsp; }&nbsp; console.log('Submitted, but something went wrong');&nbsp; console.log(response.message);}